MySQL 乘法子查询结果
我有一个看起来像的数据表
<上一页>+---------+-----------+------------+------------+|u_id |a_id |计数 |加权 |+---------+-----------+------------+------------+|1 |1 |17 |0.0521472 ||1 |2 |80 |0.245399 ||1 |3 |2 |0.00613497 ||1 |4 |1 |0.00306748 ||1 |5 |1 |0.00306748 ||1 |6 |20 |0.0613497 ||1 |7 |3 |0.00920245 ||1 |8 |100 |0.306748 ||1 |9 |100 |0.306748 ||1 |10 |2 |0.00613497 ||2 |1 |1 |0.00327869 ||2 |2 |1 |0.00327869 ||2 |3 |100 |0.327869 ||2 |4 |200 |0.655738 ||2 |5 |1 |0.00327869 ||2 |6 |1 |0.00327869 ||2 |7 |0 |0 ||2 |8 |0 |0 ||2 |9 |0 |0 ||2 |10 |1 |0.00327869 ||3 |1 |15 |0.172414 ||3 |2 |40 |0.45977 ||3 |3 |0 |0 ||3 |4 |0 |0 ||3 |5 |0 |0 ||3 |6 |10 |0.114943 ||3 |7 |1 |0.0114943 ||3 |8 |20 |0.229885 ||3 |9 |0 |0 ||3 |10 |1 |0.0114943 |+---------+-----------+------------+------------+可以用
重新创建<上一页>CREATE TABLE IF NOT EXISTS tablename (u_id INT NOT NULL, a_id MEDIUMINT NOT NULL,s_count MEDIUMINT NOT NULL, weighted FLOAT NOT NULL)ENGINE=INNODB;INSERT INTO tablename (u_id,a_id,s_count,weighted ) VALUES (1,1,17,0.0521472392638),(1,2,80,0.245398773006),(1,3,2,0.00613496932515),(1,4,1,0.00306748466258),(1,5,1,0.00306748466258),(1,6,20,0.0613496932515),(1,7,3,0.00920245398773),(1,8,100,0.306748466258),(1,4,8,460),(1,4,8,460),(1,4,8,460)1,10,2,0.00613496932515),(2,1,1,0.00327868852459),(2,2,1,0.00327868852459),(2,3,100,0.327868852459),(2,4,200,0.655737704918),(1,0.00327868852459),(2,6,1,0.00327868852459),(2,7,0,0.0),(2,8,0,0.0),(2,9,0,0.0),(2,10,1,0.00327868852459),(3,1,15,0.172413793103),(3,2,40,0.459770114943),(3,3,0,0.0),(3,4,0,0.0),(3,5,0,0.0),(3,6,10,0.114942528736),(3,7,1,0.0114942528736),(3,8,20,0.229885057471),(3,9,0,0.0),(3,10,1,0.0114942528736);我想要做的简单版本是
SELECT u_id, SUM(weighted) as total FROM tablename WHERE a_id IN (1,2,3,4,5,6,7,8,9) GROUP BY u_id ORDER BY total DESC;
给出结果
<上一页>+---------+--------------------+|u_id |总计|+---------+--------------------+|2 |0.996721301227808 ||1 |0.993865059688687 ||3 |0.988505747169256 |+---------+--------------------+我想做的更复杂的版本是根据来自 u_id 的计数对结果进行加权，因此从
中获取结果<上一页>查询 1SELECT count FROM tablename WHERE u_id = 1会返回
<上一页>+-----------+------------+|a_id |计数 |+-----------+------------+|1 |17 ||2 |80 ||3 |2 ||4 |1 ||5 |1 ||6 |20 ||7 |3 ||8 |100 ||9 |100 ||10 |2 |+-----------+------------+然后将用于计算总和，应该给出
+---------+-------------------+|u_id |总计|+---------+--------------------+|1 |83.15337423 ||3 |65.05747126 ||2 |1.704918033 |+---------+--------------------+例如用 u_id =3 计算将由
sum(查询 1 的计数值 * u_id 的加权值 = 每个 a_id 的 3)<上一页>17 * 0.172413793 =2.93103448380 * 0.459770115 =36.78160922 * 0 =01 * 0 =01 * 0 =020 * 0.114942529 =2.2988505753 * 0.011494253 =0.034482759100 * 0.229885057 =22.98850575100 * 0 =02 * 0.011494253 =0.022988506总计为 65.05747126
我怎样才能用一个查询来做到这一点?
解决方案您可以使用子查询来做到这一点.获取特定 id 计数的查询是:
SELECT a_id, s_count FROM tablename WHERE u_id = <id>您将希望将此子查询的结果左连接到主表中，然后在适当的乘法上进行 sub ，如下所示:
SELECT u_id, SUM(counts.s_count * tablename.weighted) AS total FROM tablenameLEFT JOIN (SELECT a_id, s_count FROM tablename WHERE u_id = 1) 计数ON tablename.a_id = counts.a_id按 u_id 分组I have a table of data that looks like
+---------+-----------+------------+------------+ | u_id | a_id | count | weighted | +---------+-----------+------------+------------+ | 1 | 1 | 17 | 0.0521472 | | 1 | 2 | 80 | 0.245399 | | 1 | 3 | 2 | 0.00613497 | | 1 | 4 | 1 | 0.00306748 | | 1 | 5 | 1 | 0.00306748 | | 1 | 6 | 20 | 0.0613497 | | 1 | 7 | 3 | 0.00920245 | | 1 | 8 | 100 | 0.306748 | | 1 | 9 | 100 | 0.306748 | | 1 | 10 | 2 | 0.00613497 | | 2 | 1 | 1 | 0.00327869 | | 2 | 2 | 1 | 0.00327869 | | 2 | 3 | 100 | 0.327869 | | 2 | 4 | 200 | 0.655738 | | 2 | 5 | 1 | 0.00327869 | | 2 | 6 | 1 | 0.00327869 | | 2 | 7 | 0 | 0 | | 2 | 8 | 0 | 0 | | 2 | 9 | 0 | 0 | | 2 | 10 | 1 | 0.00327869 | | 3 | 1 | 15 | 0.172414 | | 3 | 2 | 40 | 0.45977 | | 3 | 3 | 0 | 0 | | 3 | 4 | 0 | 0 | | 3 | 5 | 0 | 0 | | 3 | 6 | 10 | 0.114943 | | 3 | 7 | 1 | 0.0114943 | | 3 | 8 | 20 | 0.229885 | | 3 | 9 | 0 | 0 | | 3 | 10 | 1 | 0.0114943 | +---------+-----------+------------+------------+
which can be recreated with
CREATE TABLE IF NOT EXISTS tablename ( u_id INT NOT NULL, a_id MEDIUMINT NOT NULL,s_count MEDIUMINT NOT NULL, weighted FLOAT NOT NULL)ENGINE=INNODB; INSERT INTO tablename (u_id,a_id,s_count,weighted ) VALUES (1,1,17,0.0521472392638),(1,2,80,0.245398773006),(1,3,2,0.00613496932515),(1,4,1,0.00306748466258),(1,5,1,0.00306748466258),(1,6,20,0.0613496932515),(1,7,3,0.00920245398773),(1,8,100,0.306748466258),(1,9,100,0.306748466258),(1,10,2,0.00613496932515),(2,1,1,0.00327868852459),(2,2,1,0.00327868852459),(2,3,100,0.327868852459),(2,4,200,0.655737704918),(2,5,1,0.00327868852459),(2,6,1,0.00327868852459),(2,7,0,0.0),(2,8,0,0.0),(2,9,0,0.0),(2,10,1,0.00327868852459),(3,1,15,0.172413793103),(3,2,40,0.459770114943),(3,3,0,0.0),(3,4,0,0.0),(3,5,0,0.0),(3,6,10,0.114942528736),(3,7,1,0.0114942528736),(3,8,20,0.229885057471),(3,9,0,0.0),(3,10,1,0.0114942528736);
the simple version of what I want to do is
SELECT u_id, SUM(weighted) as total FROM tablename WHERE a_id IN (1,2,3,4,5,6,7,8,9) GROUP BY u_id ORDER BY total DESC;
which gives the results
+---------+-------------------+ | u_id | total | +---------+-------------------+ | 2 | 0.996721301227808 | | 1 | 0.993865059688687 | | 3 | 0.988505747169256 | +---------+-------------------+
the more complex version I want to do is to weight the results based on the count from a u_id, so taking the results from
query 1 SELECT count FROM tablename WHERE u_id = 1
would return
+-----------+------------+ | a_id | count | +-----------+------------+ | 1 | 17 | | 2 | 80 | | 3 | 2 | | 4 | 1 | | 5 | 1 | | 6 | 20 | | 7 | 3 | | 8 | 100 | | 9 | 100 | | 10 | 2 | +-----------+------------+
which would then be used to calculate the sum, should give
+---------+-------------------+
| u_id | total |
+---------+-------------------+
| 1 | 83.15337423 |
| 3 | 65.05747126 |
| 2 | 1.704918033 |
+---------+-------------------+
eg calculating with u_id =3 would be done by
sum(count value from query 1 * weighting value for u_id = 3 for each a_id)
17 * 0.172413793 =2.931034483 80 * 0.459770115 =36.7816092 2 * 0 =0 1 * 0 =0 1 * 0 =0 20 * 0.114942529 =2.298850575 3 * 0.011494253 =0.034482759 100 * 0.229885057 =22.98850575 100 * 0 =0 2 * 0.011494253 =0.022988506 sums up to 65.05747126
How can I do this with a single query?
解决方案You can do this using a subquery. The query that gets the counts for a particular id is:
SELECT a_id, s_count FROM tablename WHERE u_id = <id>
You will want to left join the result of this subquery into the main table, then sub over the appropriate multiplication, like so:
SELECT u_id, SUM(counts.s_count * tablename.weighted) AS total FROM tablename
LEFT JOIN (SELECT a_id, s_count FROM tablename WHERE u_id = 1) counts
ON tablename.a_id = counts.a_id
GROUP BY u_id
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